BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2075 Nepal

Definition of the limit of a function

A function $f(x)$ is said to have the limit $L$ as $x \to a$, written $\lim_{x\to a} f(x) = L$, if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

Intuitively, $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $a$ (but not equal to $a$).

Evaluating $\lim_{x\to 0}\dfrac{1-\cos x}{x^2}$

This is of the indeterminate form $\tfrac{0}{0}$. Using the identity $1 - \cos x = 2\sin^2\!\frac{x}{2}$:

Since $\lim_{u\to 0}\dfrac{\sin u}{u} = 1$ (with $u = x/2$),

(Equivalently, by L'Hôpital's rule applied twice: $\frac{\sin x}{2x} \to \frac{\cos x}{2} = \frac12$.)

Continuity

The function $g(x) = \dfrac{1-\cos x}{x^2}$ is not defined at $x = 0$ (denominator vanishes), so it has a removable discontinuity there. Defining

makes $g$ continuous everywhere, because $\lim_{x\to 0} g(x) = \tfrac12 = g(0)$. For $x \neq 0$, $g$ is a quotient of continuous functions with non-zero denominator, hence continuous.

Leibnitz's Theorem

If $u$ and $v$ are functions of $x$ each possessing derivatives up to the $n$th order, then the $n$th derivative of their product is

where $u_k = \dfrac{d^k u}{dx^k}$ and $v_k = \dfrac{d^k v}{dx^k}$.

$n$th derivative of $y = \sin^{-1}x$ at $x = 0$

Step 1 — Form a differential equation. Let $y = \sin^{-1}x$. Then

Differentiating: $(1-x^2)\,2y_1 y_2 - 2x\,y_1^2 = 0$, i.e.

Step 2 — Apply Leibnitz's theorem. Differentiate $(1-x^2)y_2 - x y_1 = 0$ $n$ times. Using Leibnitz on each product:

which simplifies to the recurrence

Step 3 — Put $x = 0$. Writing $y_n(0) = (y_n)_0$,

Initial values: $y(0) = 0$, $y_1(0) = 1$, $y_2(0) = 0$.

Step 4 — Conclusion.

• If $n$ is even, $(y_n)_0 = 0$.
• If $n$ is odd, say $n = 2m+1$, then iterating the recurrence,

Thus

e.g. $y_3(0) = 1^2 = 1$, $y_5(0) = 1^2\cdot 3^2 = 9$, $y_7(0) = 1^2\cdot 3^2\cdot 5^2 = 225$.

Volume of revolution about the x-axis

For a region bounded by $y = f(x)$, $x = a$ and $x = b$ revolved about the x-axis, the disk method gives

Given: $y = x^2$, $x = 0$ to $x = 2$, revolved about the x-axis.

Integrate:

This is the standard limit defining the number $e$.

Let $L = \lim_{x\to\infty}\left(1 + \dfrac{1}{x}\right)^x$. Taking logarithms,

Put $t = 1/x \to 0$: $\ln L = \lim_{t\to 0}\dfrac{\ln(1+t)}{t} = 1$ (using $\ln(1+t) = t - \tfrac{t^2}{2}+\cdots$, or L'Hôpital).

Hence $\ln L = 1 \Rightarrow L = e$.

Let $y = x^x$. Take logarithms (logarithmic differentiation):

Differentiate both sides with respect to $x$:

Therefore

Lagrange's Mean Value Theorem (statement)

If $f(x)$ is (i) continuous on the closed interval $[a,b]$ and (ii) differentiable on the open interval $(a,b)$, then there exists at least one point $c \in (a,b)$ such that

Verification for $f(x) = x^2$ on $[2,4]$

Hypotheses: $f(x) = x^2$ is a polynomial, hence continuous on $[2,4]$ and differentiable on $(2,4)$. The conditions hold.

Compute the average rate of change:

Solve $f'(c) = 6$: Since $f'(x) = 2x$,

Since $c = 3 \in (2,4)$, the theorem is verified.

Let $f(x) = x^3 - 3x + 2$.

Step 1 — Critical points. $f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 0 \Rightarrow x = \pm 1.$

Step 2 — Second-derivative test. $f''(x) = 6x$.

• At $x = -1$: $f''(-1) = -6 < 0 \Rightarrow$ local maximum.
• At $x = 1$: $f''(1) = 6 > 0 \Rightarrow$ local minimum.

Step 3 — Values.

Conclusion: Local maximum value $= 4$ at $x = -1$; local minimum value $= 0$ at $x = 1$. (Note: being a cubic, $f$ has no global maximum or minimum on $\mathbb{R}$.)

Resolve $\dfrac{x}{(x+1)(x+2)}$ into partial fractions:

Then $x = A(x+2) + B(x+1)$.

• Put $x = -1$: $-1 = A(1) \Rightarrow A = -1.$
• Put $x = -2$: $-2 = B(-1) \Rightarrow B = 2.$

So

Integrate:

The equation $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$ is homogeneous (numerator and denominator are degree-1 in $x,y$).

Substitute $y = vx \Rightarrow \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}.$ Then

So

Separate variables:

Integrate the left side:

Thus

Back-substitute $v = y/x$, and use $1 + v^2 = \dfrac{x^2+y^2}{x^2}$:

which simplifies (since $\tfrac12\ln\frac{x^2+y^2}{x^2} = \tfrac12\ln(x^2+y^2) - \ln x$) to

A vector perpendicular to both $\vec a$ and $\vec b$ is given by the cross product $\vec a \times \vec b$.

Compute $\vec a \times \vec b$ with $\vec a = 2\hat i + \hat j + \hat k$, $\vec b = \hat i - \hat j + 2\hat k$:

• $\hat i$: $(1)(2) - (1)(-1) = 2 + 1 = 3$
• $\hat j$: $-\big[(2)(2) - (1)(1)\big] = -(4 - 1) = -3$
• $\hat k$: $(2)(-1) - (1)(1) = -2 - 1 = -3$

So $\vec a \times \vec b = 3\hat i - 3\hat j - 3\hat k$.

Magnitude: $|\vec a \times \vec b| = \sqrt{3^2 + (-3)^2 + (-3)^2} = \sqrt{27} = 3\sqrt3.$

Unit vector:

Given $z = x^2 y + x y^2$.

First-order partials:

Mixed partial $\dfrac{\partial^2 z}{\partial y\,\partial x}$ (differentiate $\partial z/\partial x$ w.r.t. $y$):

Mixed partial $\dfrac{\partial^2 z}{\partial x\,\partial y}$ (differentiate $\partial z/\partial y$ w.r.t. $x$):

Since both equal $2x + 2y$,

verifying the equality of mixed partial derivatives (Clairaut/Young's theorem), as expected since $z$ has continuous second-order partials.

Evaluate the iterated integral, integrating with respect to $y$ first (for $0 \le y \le x$):

Inner integral (treat $x$ as constant):

Outer integral: